# A Bit of Physics: Trajectories and Projectile Motion

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In my A-level Physics classes I’ve come across the function that plots the trajectory of a projectile launched at an angle $θ$ with initial velocity $v$.$$y=xtanθ-\tfrac{g}{2v^2cosθ}$$

I have been trying to understand this equation intuitively for around two weeks without success, and in this topic I’m going to share my insights so you guys can maybe help me out. I’m going to start with the basics in order not to exclude people without any background in physics from the conversation.

Just like math, physics has many sub-branches, and kinematics describes how objects move without considering what caused them to.
For example it might tell you that you are falling from a 10 meter trampoline with a constant acceleration, and that in .5 seconds you are going to feel the worse belly flop of your entire life, but it won’t ever tell you if it was that douchebag Chad who pushed you to your doom or if you actually slipped on that soap bar which had been left there by an unsuspecting lifeguard.

URM and URAM
Let’s suppose you’re Angus Young. You’re “on [your] way to the promised land, [you’re] on the higway to hell”.
However, even though there was “no speed limit” until recently, Satan (or “the adversary” as Ben Carson calls him. If you don’t know what I mean, consider yourself a lucky person) has introduced one, just to make hell more hellish.
So you’re late for your appointment with your friends who “are gonna be there too”, but you’re already travelling at the maximum speed. You do a quick calculation in your head and realise that if you’re going to continue at this speed and go for the Uniform Rectilinear Motion you’re not going to make it. Therefore you decide to go flat out and stack your hopes on the Uniformely Accelerated Rectilinear Motion, at the risk of getting a ticket (who am I kidding? If anything they’re going to praise you!).

Now you may be asking yourself: what on earth is this guy talking about? Just watch:$$s=vt$$What this equation is saying is that the space you travel is equal to the rate (how many units of space you cover in one unit of time, aka the constant velocity) times the span of time during which you maintain this velocity.
But what if, like Angus Young, you feel like you’re going too slow? In that case acceleration comes into play, and you have:$$s=v_0t+\tfrac{1}{2}at^2$$Notice that the first formula is just a specific scenario of this one: namely, when there is no acceleration and thus $a=0$ (keep in mind that for this formula to work acceleration must be constant). $v_0$ stands for the initial velocity, but why is it $\tfrac{1}{2}at^2$? The short answer is that the $\tfrac{1}{2}$ stands for the fact that you need the average speed to calculate the acceleration (which is calculated by subtracting your initial speed from how fast you are going after a certain period of time, or $a=\tfrac{v-v_0}{t}$) because it’s constantly changing, and $t^2$ stands for the fact that the time during which you maintain the acceleration influences both the space covered and the final velocity, which in turn influences the space covered. The longer answer will be in the appendix.

Independence of Perpendicular Components of Motion
Let’s say we want to apply the previous formulas to a plane with two dimensions. The first thing that comes to mind is an x-y coordinate plane, but immediately a problem emerges: the formulas we have are apparently only useful for one-dimensional motion!
Fortunately Galileo has already solved this problem for us. He discovered that motion in one dimension doesn’t affect motion in a dimension perpendicular to it. This makes sense: if you draw an x-y graph and move along the x-axis, your position on the y-axis remains 0. This however doesn’t work if the angle between the two axes is bigger or smaller than $90^o$ (try it!), and this is why they must be perpendicular.
When I was in my freshman year in middle-school I had an awful math teacher who would make us memorise definitions without even understanding them and make us repeat them over and over like parrots. Whenever I asked her something, like “why are the axes always orthogonal?” (orthogonal is a fancy word meaning perpendicular, although at the time I didn’t know as I had just learnt it by rote) she would just say: “it doesn’t matter right now, just make sure to memorise what I told you to”. She almost made me quit math. Well, now you know why Cartesian coordinates are like they are.

Projectile Motion
Finally we come to projectile motion. first of all what is a projectile? It’s simply a name we give to objects which have an internal structure that doesn’t influence their motion (for example an egg isn’t a projectile) and whose trajectory is influenced exclusively by the force of gravity. Some examples are cannonballs, people falling from a trampoline and archer fish sharpshooting insects.
To describe such motion we break it down into two dimensions (if the system we were analysing had three dimensions we would have to act accordingly, but most things can be analysed by using a 2d perspective) along the x-axis, there is no force in action, so we write $s_x=v_xt$ along the y-direction, however, there is an acceleration due to gravity, which we will call $g=9,81\tfrac{m}{s^2}$ (the acceleration is measured in $\tfrac{m}{s^2}$ because for every second, the speed of a falling object augments by $9,81\tfrac{m}{s}$) so we write $s_y=v_{0y}t-\tfrac{1}{2}gt^2$ and the minus in front of $g$ stands for the fact that objects when falling usually tend to go in the downward direction.
To get the path of an object we have to remove the $t$ variable, because when plotting the trajetory we are not interested in the object which is falling but only in the function that describes its motion. Furthermore, since we want an equation in the form $y=f(x)$ we extract $t$ from the equation $s_x=v_xt$ and replace it in the equation $s_y=v_{0y}t-\tfrac{1}{2}gt^2$ and we get something like$$s_y=v_{0y}(\tfrac{s_x}{v_x})-\tfrac{1}{2}g(\tfrac{s_x}{v_x})=\tfrac{v_{0y}}{v_x}x-(\tfrac{g}{2v_x^2})x^2$$which is an equation in the form $y=ax+bx^2$ with $a=\tfrac{v_{0y}}{v_x}$ and $b=-\tfrac{g}{2v_x^2}$ which are both constant. We notice with awe that it’s the equation of a parabola with the origin intersecting with the axes’ origin (where we throw the object from).
By imagining $v$ as a vector we can call the angle at which this vector is applied $θ$ and we get $$y=xtanθ-\tfrac{g}{2v^2cosθ}$$ since the x component of a vetor can be calculated with trig.
Now we know how to get the equation algebraically, but we still don’t understand it intuitively. I asked my physics teacher for advice without success, so I started digging on my own. Here are my notes:

• Wikipedia says: "Assume the motion of the projectile is being measured from a Free fall frame which happens to be at $(x,y)=(0,0)$ at $t=0$. The equation of motion of the projectile in this frame (by the principle of equivalence) would be $y = x \tan(\theta)$. The co-ordinates of this free-fall frame, with respect to our inertial frame would be $y = - gt^2/2$. That is, $y = - g(x/v_h)^2/2$.

Now translating back to the inertial frame the co-ordinates of the projectile becomes $y = x \tan(\theta)- g(x/v_h)^2/2$ That is:

$y=-{g\sec^2\theta\over 2v_0^2}x^2+x\tan\theta$,

(where $v_0$ is the initial velocity, $\theta$ is the angle of elevation, and $g$ is the acceleration due to gravity).$Appendix We know that the equation for the Uniform Rectilinear Motion is $$s=vt$$ and that velocity is constantly changing in the Uniformely Accelerated Rectilinear Motion. A way to bypass that (besides calculus) is to take average velocity$\tfrac{v_0+v}{2}$and multiply it by the time to get$s=\tfrac{v_0+v}{2}t$. If we know the acceleration and we want to know the final velocity we use the equation$v=v_0+at$(which is just an inverse formula of$a=\tfrac{v-v_0}{t}$) and substitute it in our equation for space to get$s=\tfrac{v_0+(v_0+at)}{2}t=v_0t+\tfrac{1}{2}at^2\$